If empirical formula of compound 3 is $P_{3} Q_{4}$, then the empirical formula of compound 2 is $\mathrm{P}_{3} \mathrm{Q}_{5}$.

If empirical formula of compound 3 is $\mathrm{P}_{3} \mathrm{Q}_{2}$ and atomic weight of element $\mathrm{P}$ is 20 , then the atomic weight of $\mathrm{Q}$ is 45 .

If empirical formula of compound 2 is $P Q$, then the empirical formula of the compound $\mathbf{1}$ is $\mathrm{P}_{5} \mathrm{Q}_{4}$.

If atomic weight of $\mathrm{P}$ and $\mathrm{Q}$ are 70 and 35 , respectively, then the empirical formula of compound $\mathbf{1}$ is $\mathrm{P}_{2} \mathrm{Q}$.

For the reaction, $\mathrm{M}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{H}_{2}(g)+\mathrm{M}^{2+}(a q)$, if $\frac{d E_\text{cell}}{d T}=\frac{R}{F}$, then the entropy change of the reaction is $\mathrm{R}$ (assume that entropy and internal energy changes are temperature independent).

The cell reaction, $\operatorname{Pt}(s) \mid \mathrm{H}_{2}(g, 1$ bar $)\left|\mathrm{H}^{+}(a q, 0.01 \mathrm{M}) \| \mathrm{H}^{+}(a q, 0.1 \mathrm{M})\right| \mathrm{H}_{2}(g, 1 \mathrm{bar}) \mid \operatorname{Pt}(s)$, is an entropy driven process.

For racemization of an optically active compound, $\Delta \mathrm{S}>0$.

$\Delta \mathrm{S}>0$, for $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+3$ en $\rightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}+6 \mathrm{H}_{2} \mathrm{O}$ (where en $=$ ethylenediamine).

$\mathrm{B}$

$\mathrm{B}_{2} \mathrm{H}_{6}$

$\mathrm{B}_{2} \mathrm{O}_{3}$

$\mathrm{HBF}_{4}$